AP Physics C Electricity and Magnetism Quiz
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ELECTRIC POTENTIAL
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A 12 V12\,\text{V} ideal battery charges a C=200 μFC=200\,\mu\text{F} capacitor through a resistor R=300 ΩR=300\,\Omega; after a long time, the capacitor reaches 12 V12\,\text{V}. The capacitor then disconnects from the battery and discharges through the same resistor in a closed loop. During discharge, charge moves through a potential difference, so capacitor potential energy decreases and becomes thermal energy in RR (via brief carrier kinetic energy). The initial discharge energy is U0=12CV2=12(200×10−6)(122)=1.44×10−2 JU_0=\tfrac12 C V^2=\tfrac12(200\times10^{-6})(12^2)=1.44\times10^{-2}\,\text{J}. With RC=0.060 sRC=0.060\,\text{s}, after t=0.030 st=0.030\,\text{s}, UC=U0e−2t/RC=U0e−1≈5.30×10−3 JU_C=U_0 e^{-2t/RC}=U_0 e^{-1}\approx5.30\times10^{-3}\,\text{J}, so 9.1×10−3 J9.1\times10^{-3}\,\text{J} has been converted to heat while total energy remains conserved. Based on the scenario, how does the energy stored in the capacitor change as it discharges?
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